1. Tardigrade
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  4. A seconds pendulum clock has a steel wire. The clock shows correct time at 25° C. How much time does the clock lose or gain, in one week, when the temperature is increased to 35° C ? (α text steel =1.2 × 10-5 / ° C )

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Q. A seconds pendulum clock has a steel wire. The clock shows correct time at $25^{\circ} C$. How much time does the clock lose or gain, in one week, when the temperature is increased to $35^{\circ} C$ ?
$\left(\alpha_{\text {steel }}=1.2 \times 10^{-5} /{ }^{\circ} C \right)$

Thermal Properties of Matter

Solution:

$\frac{\Delta T}{T}=\frac{1}{2} \,\alpha \,\theta$
$=\frac{1}{2} \times 1.2 \times 10^{-5} \times 10$
$\frac{\Delta T}{T}=6.0 \times 10^{-5}$
Hence time lost in $1$ week $=6.0 \times 10^{-5} \times T$
$=6.0 \times 10^{-5} \times 7 \times 24 \times 3600$
$=36.28\, s$